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Vector $\lambda \overrightarrow{a}$ is a unit vector if $\left|\lambda \overrightarrow{a}\right|=1$

Now,

$\begin{array}{l}\left|\lambda \overrightarrow{a}\right|=1\\ \left|\lambda \right|\left|\overrightarrow{a}\right|=1\\ \left|\overrightarrow{a}\right|=\frac{1}{\left|\lambda \right|}\left[\lambda \ne 0\right]\\ a=\frac{1}{\left|\lambda \right|}\left[\left|\overrightarrow{a}\right|=a\right]\end{array}$

$\therefore$ Therefore, vectar $\lambda \overrightarrow{a}$ is a unit vector if a= $\frac{1}{\left|\lambda \right|}$ .

Option (D)is correct.

Saveetha College of Architecture and Design admissions are mix of entrance or merit-based as per course requirement. Candidates can register online for admission to their preferred course. The candidate must meet the eligibility criteria set by the college. Aspirants must pass Class 12 for the UG courses and graduation for the PG courses. The college also considers the candidates' academic performance for final selection.

Let vector $2\widehat{i}-\widehat{j}+\widehat{k},\widehat{i}-3\widehat{j}-5\widehat{k}$ and $3\widehat{i}-4\widehat{j}-4\widehat{k}$ be position vector of point A, B, C respectively.

So,

$\begin{array}{l}O\overrightarrow{A}=2\widehat{i}-\widehat{j}+\widehat{k}\\ O\overrightarrow{B}=\widehat{i}-3\widehat{j}-5\widehat{k}\\ O\overrightarrow{C}=3\widehat{i}-4\widehat{j}-4\widehat{k}\end{array}$

Now, vectors $A\overrightarrow{B},B\overrightarrow{C}$ and $A\overrightarrow{C}$ represents the sides of $?ABC$ .

Hence,

Given, point are

Vertices of $?ABC$ are given as

$\begin{array}{l}A\left(1,2,3\right),B\left(-1,0,0\right),C\left(0,1,2\right)\\ \end{array}$

$\therefore ?ABC$ is the angle between the vectors $B\overrightarrow{A}$ and $B\overrightarrow{C}$

Consider

$\begin{array}{l}\overrightarrow{a}=2\widehat{i}+4\widehat{j}+3\widehat{k}\\ \overrightarrow{b}=3\widehat{i}+3\widehat{j}-6\widehat{k}\end{array}$ and

Then,

$\begin{array}{l}\overrightarrow{a}.\overrightarrow{b}=2.3+4.3+3.(-6)\\ =6+12-18=0\end{array}$

Therefore, the converse of the given statement need not be true.

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Additionally, the university offers BCA specializations in Data Analytics and Machine Learning, but not specifically in Cyber Security.

SCAD offers BArch, MArch, BSc, etc., at the UG and PG levels. The selection criteria are merit or entrance-based.The institute offers various courses in the stream of Architecture and Design. The mode of study of the programme is full-time. Students can check the table below to know course-wise total tuition fee at Saveetha College of Architecture and Design:

Note: This information is sourced from official website/ sanctioning body and is subject to change.

Candidates who wish to apply for the S. S. Jain Subodh Management Institute MBA course must ensure that they fulfil the eligibility criteria. As per the institute's official website, any candidate who has completed a UG degree in any stream with a minimum of 50% in aggregate is eligible to apply. Moreover, students of reserved categories need only 45% in graduation to be eligible.

$\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right).\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$

$\begin{array}{l}=\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}\\ ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)\\ =1+1+1+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)\\ =3+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)\\ \Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=\frac{-3}{2}\end{array}$

50.The given system of inequality is

3x+2y≤ 150- (1)

x+4y≤ 80- (2)

x≤ 15 - (3)

y≥ 0 and x≥ 0 - (4)

The corresponding equation of (1) and (2) are

3x + 2y = 150

and x + 4y =80

Putting (0,0)= (x, y) in inequality (1) and (2) we get,

3 × 0+2 × 0 ≤ 150  => 0 ≤ 150 is true.

and 0+4 × 0 ≤ 80  => 0 ≤ 80 is true.

So, the solution plane of both inequality (1) and (2) includes the origin (0,0).

∴ The shaded region is the solution of the given system of inequality.

We know,

$\overrightarrow{a}.\overrightarrow{a}=0$ and $\overrightarrow{a}.\overrightarrow{b}=0$

Now,

$\overrightarrow{a}.\overrightarrow{a}=0\Rightarrow {\left|\overrightarrow{a}\right|}^2\Rightarrow \left|\overrightarrow{a}\right|=0$

$\therefore$ $\overrightarrow{a}$ is a zero vector.

Thus, vector $\overrightarrow{b}$ satisfying $\overrightarrow{a}.\overrightarrow{b}=0$ can be any vector.

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Disclaimer: This information is sourced from official website / media reports/ NIRF website and may vary.

2. Given, f (x) = 2x2 1

At x = 3

Lim f (x) = $\frac{dy}{dx}=\frac{-\left(3x^2+2xy+y^2\right)}{\left(x^2+2xy+3y^2\right).}$ 2 (3)2 1 = 18 1 = 17.

So, f is continuous at x = 3.

$\left(\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}\right).\left(\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}\right)$

$\begin{array}{l}=\left|\overrightarrow{a}\right|\overrightarrow{b}.\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{a}\right|\overrightarrow{b}.\left|\overrightarrow{b}\right|\overrightarrow{a}+\left|\overrightarrow{b}\right|\overrightarrow{a}.\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}.\left|\overrightarrow{b}\right|\overrightarrow{a}\\ ={\left|\overrightarrow{a}\right|}^2\overrightarrow{b}.\overrightarrow{b}-{\left|\overrightarrow{b}\right|}^2\overrightarrow{a}.\overrightarrow{a}\\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{b}\right|}^2{\left|\overrightarrow{a}\right|}^2\\ =0\end{array}$

$\therefore$ Therefore, $\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}$ and $\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}$ are perpendicular.

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Disclaimer: This information is sourced from official website.